Find the largest palindrome made from the product of two 3-digit numbers

package testing.project;

public class PalindromeThreeDigits {

    public static void main(String[] args) {
        int value = 0;
        for(int i = 100;i <=999;i++)
        {
            for(int j = i;j <=999;j++)
            {
                int value1 = i * j;
                StringBuilder sb1 = new StringBuilder(""+value1);
                String sb2 = ""+value1;
                sb1.reverse();
                if(sb2.equals(sb1.toString()) && value<value1) {
                    value = value1;

                }

            }
        }

        System.out.println(value);
    }
}

This is the code that I wrote in Java... Is there any efficient way other than this.. And can we optimize this code more??

We suppose the largest such palindrome will have six digits rather than five, because 143*777 = 111111 is a palindrome.

As noted elsewhere, a 6-digit base-10 palindrome abccba is a multiple of 11. This is true because a*100001 + b*010010 + c*001100 is equal to 11*a*9091 + 11*b*910 + 11*c*100. So, in our inner loop we can decrease n by steps of 11 if m is not a multiple of 11.

We are trying to find the largest palindrome under a million that is a product of two 3-digit numbers. To find a large result, we try large divisors first:

• We step m downwards from 999, by 1's;
• Run n down from 999 by 1's (if 11 divides m, or 9% of the time) or from 990 by 11's (if 11 doesn't divide m, or 91% of the time).

We keep track of the largest palindrome found so far in variable q. Suppose q = r·s with r <= s. We usually have m < r <= s. We require m·n > q or n >= q/m. As larger palindromes are found, the range of n gets more restricted, for two reasons: q gets larger, m gets smaller.

The inner loop of attached program executes only 506 times, vs the ~ 810000 times the naive program used.

#include <stdlib.h>
#include <stdio.h>
int main(void) {
  enum { A=100000, B=10000, C=1000, c=100, b=10, a=1, T=10 };
  int m, n, p, q=111111, r=143, s=777;
  int nDel, nLo, nHi, inner=0, n11=(999/11)*11;

  for (m=999; m>99; --m) {
    nHi = n11;  nDel = 11;
    if (m%11==0) {
      nHi = 999;  nDel = 1;
    }
    nLo = q/m-1;
    if (nLo < m) nLo = m-1;

    for (n=nHi; n>nLo; n -= nDel) {
      ++inner;
      // Check if p = product is a palindrome
      p = m * n;
      if (p%T==p/A && (p/B)%T==(p/b)%T && (p/C)%T==(p/c)%T) {
        q=p; r=m; s=n;
        printf ("%d at %d * %d\n", q, r, s);
        break;      // We're done with this value of m
      }
    }
  }
  printf ("Final result:  %d at %d * %d   inner=%d\n", q, r, s, inner);
  return 0;
}

Note, the program is in C but same techniques will work in Java.